To The Max

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5664    Accepted Submission(s): 2686

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

is in the lower left corner:

9 2
-4 1
-1 8

and has a sum of 15.

 

Input

The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].

 

Output

Output the sum of the maximal sub-rectangle.

 

Sample Input

4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2

 

Sample Output

15

题目没明说但必须考虑到有多组数据（用while）

#include 
#include 
#include 
#include 
#include 
using namespace std;
int num[110][110];
int sum[110][110];
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        int i,j;
        for(i=1;i<=n;++i)
          for(j=1;j<=n;++j)
            scanf("%d",&num[i][j]);
        for(i = 0;i<=n;++i)
        {
            sum[0][i] = 0;
            sum[i][0] = 0;
        }
        for(i = 1;i<=n;++i)
           for(j = 1;j<=n;++j)
             sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + num[i][j];
        int maxn = num[1][1];
        for(i = n;i>=1;--i)
           for(j = n;j>=1;--j)
             for(int k = 0;k<=i;++k)
               for(int l = 0;l<=j;++l)
                  {
                      int temp = sum[i][j] - sum[k][j] - sum[i][l] +sum[k][l];
                      maxn = max(temp,maxn);
                  }
         printf("%d\n",maxn);
    }
    return 0;

}